95% C.I. estimate

[LondonBroil]

Wifey has a co-worker who either has a homework problem/take-home test/whatever and she can't figure out this problem. I'm taking a prob/stat course now but we haven't gotten to this yet. Anyone?

The problem is :



2 independent samples are randomly selected from 2 normally distributed populations

Sample 1: 22,16,15,18,21,15 Mean 17.83 S 3.06
Sample 2: 17,21,23,15,18,17 Mean 18.50 S 2.95
Difference 5,-5,-8,3,3,-2 -.67 S 5.16

The 95% C.I. est for the difference in the population mean is:

[LondonBroil]

I guess she needs to show her work:



Hi Matt,

If this helps the answers are below but I tried using my formulas and was not able to come up with the answer
(-4.6,3.3) (-4.5,3.2) (-6.09, 4.75)

[DiceyPlay]

I would use a 2 sample t test here

95% CI = (mean1 - mean2) +- t-sub(.025, df=5)*sqrt(s1^2/n1+s2^2/n2) = -.67 +- 2.5706 * 1.74 = (-5.13, 3.79).

But I think the difference can not be negative, so I would opt for (0, 5.13) as a 95% CI.

What am I doing wrong or thinking wrongly?


This isn't any of the answers you suggested.

[jman3232]

1) the difference can be negative, it would just mean that mean2 is bigger than mean1.
2) this is not a ttest, it should be a 2sample t interval.